This may be more involved than necessary, but I'm using this as an exercise to make sure I'm thinking of this correctly. Say we have:
$$f(x) = \frac{2x^2 - x + 1}{x^2 +4}$$
And we want to show that the limit as $x \to \infty$ is $2$. I'll need to show that $\forall \epsilon > 0 ~~ \exists ~\delta : x > \delta \implies |f(x) - 2| \leq \epsilon . $
Does it suffice to say:
\begin{align} \left |\frac{2x^2 - x + 1}{x^2 + 4} - 2 \right | &\leq \epsilon &\implies \\ \\ \frac{2x^2 - x + 1}{x^2 + 4} - 2 &\leq \epsilon &\implies \\ \\ \frac{-2x^2 + x - 1}{x^2 + 4} &\geq \epsilon + 2 &\implies \\ \\ -2x^2 + x - 1 & \geq \epsilon + 2 &\implies \\ \\ -2x^2 + x &\geq \epsilon + 3 &\implies \\ x &\geq \epsilon + 3 \end{align}
and
\begin{align} \left |\frac{2x^2 - x + 1}{x^2 + 4} - 2 \right | &\leq \epsilon &\implies \\\\ 2 - \frac{2x^2 - x + 1}{x^2 + 4} &\leq \epsilon &\implies \\ \\ \frac{2x^2-x+1}{x^2 + 4} &\geq \epsilon + 2 &\implies \\ \\ 2x^2 - x + 1 &\geq \epsilon + 2 &\implies \\ \\ 2x^2 + 1 &\geq \epsilon + 2 &\implies \\ \\ 2x^2 &\geq \epsilon + 1 &\implies \\ \\ x &\geq \sqrt{\frac{\epsilon + 1}{2}} \end{align}
Since $\epsilon + 3 > \sqrt{\frac{\epsilon + 1}{2}}$ when $\epsilon > 0$—if we choose $\delta = \epsilon + 3$, then $|f(x) = 2| \leq \epsilon$.
Does this look correct?
$$\frac{2x^2-x+1}{x^2+4}-2\leq \epsilon\Rightarrow \frac{-2x^2+x-1}{x^2+4}\geq \epsilon+2$$ is definitely wrong.
You may not think of an inequality but how to modify or estimate the RHS such that you see that it has to be less than $\epsilon$, using the free choice of $\delta$.
You should try it this way: \begin{align} \left|\frac{2x^2-x+1}{x^2+4}-2\right|&=\left|\frac{2x^2-x+1}{x^2+4}-\frac{2(x^2+4)}{x^2+4}\right|=\ldots\\ &=\frac{x+7}{x^2+4}\leq\ldots\leq \frac1{2x}+\frac7{2x^2}\\ &\leq \frac1{2\delta}+\frac{7}{2\delta^2}. \end{align} Now you can choose $\delta$ such that it is less than $\epsilon$.